Find the derivative of y with respect to x.
-$$y = \tan ^ { - 1 } ( \ln 4 x )$$
A) $\frac { 4 } { x \left( 1 + ( \ln 4 x ) ^ { 2 } \right) }$
B) $\frac { 1 } { x \left( 1 + ( \ln 4 x ) ^ { 2 } \right) }$
C) $\frac { 1 } { 1 + ( \ln 4 x ) ^ { 2 } }$
D) $\frac { 1 } { x \sqrt { 1 + ( \ln 4 x ) ^ { 2 } } }$

Question

Quizplus · Accepted Answer

The derivative of $y = \tan^{-1}(\ln 4x)$ is found using the chain rule. The derivative of $\tan^{-1}(u)$ is $\frac{1}{1+u^2}$, and the derivative of $\ln 4x$ is $\frac{1}{x}$. Combining these, the derivative is $\frac{1}{x(1+(\ln 4x)^2)}$.

Find the Derivative of Y with Respect to X y=tan⁡−1(ln⁡4x)y = \tan ^ { - 1 } ( \ln 4 x )y=tan−1(ln4x)

Find the Derivative of Y with Respect to X $y = \tan ^ { - 1 } ( \ln 4 x )$